Generating binomial coefficients in a row of Pascal's triangle from extensions of powers of eleven

Sir Isaac Newton noticed that the values of the first five rows of Pascal's triangle are each formed by a power of 11, and claimed that subsequent rows can also be generated by a power of 11. Literally, the claim is not true for the 5th row and onward. His genius mind might have suggested a deep relation between binomial coefficients and a power of some integer that resembles the number 11 in some form. In this study, we propose and prove a general formula to generate the values in any row of Pascal's triangle from the digits of (10⋯0︸Θ zeros1)n. It can be shown that the numbers in the cells in nth row of Pascal's triangle may be achieved from Θ+1 partitions of the digits of the number (10⋯0︸Θ zeros1)n, where Θ is a non-negative integer. That is, we may generate the number in the cells in a row of Pascal's triangle from a power of 11, 101, 1001, or 10001 and so on. We briefly discuss how to determine the number of zeros Θ in relation to n, and then empirically show that the partition really gives us binomial coefficients for several values of n. We provide a formula for Θ and prove that the (n+1)th row of Pascal's triangle is simply Θ+1 partitions of the digits of (10⋯0︸Θ zeros1)n from the right.


Introduction
Algebra is a spacious part of the science of mathematics that provides the opportunity to express mathematical ideas precisely. In algebra, the binomial expansion and Pascal's triangle are considered important. Pascal's triangle is an arrangement of the binomial coefficients and one of the most known integer models. Though it was named after the French scientist Blaise Pascal, it was studied in ancient India [1,2], Persia [3,4], China [5], Germany, and Italy [6].
In reality, the definition of the triangle was made centuries ago. In 450 BC, an Indian mathematician named Pingala is said to have introduced the definition of this triangle in a Sanskrit poetry book. Chinese mathematicians had the same idea and named the triangle as "Yang Hui's triangle". Later, Persian mathematician Al-Karaji and Persian astronomer-poet Omar Khayyam named the triangle as the "Khayyam triangle". It also has multi-dimensional shapes. The three-dimensional shape is referred to as Pascal's pyramid or Pascal's tetrahedron, while the other general-shaped ones are called Pascal's simplifications.
Various studies have been conducted in many different disciplines about Pascal's triangle. For the construction of Pascal's triangle, Sgroi [7] stated that each line starts with 1 and ends with 1, and this series can be expanded with simple cross-joints. Jansson [8] developed three geometric forms related to Pascal's triangle and included examples of each form. Toschi [9] used various permutations to generate new forms of Pascal's triangles and generalized them. Duncan and Litwiller [10] addressed the reconstruction of Pascal's triangle with the individuals. Here they collected data on the opinions of individuals using qualitative methods, and determined the methods of constructing the Pascal's triangle in different ways with the attained findings.
Researchers worked on Pascal's fascinating characteristics. Using the principle of permutation, Putz [11] designed Pascal Polytope and linked it to the Fibonacci concept. Houghton [12] gave the concept of the relationship between the successive differential operations of a function and Pascal's triangle. With an application, he attempted to incorporate the idea of a differentiable function into Pascal's triangle. The relationship between Pascal's triangle and the Tower of Hanoi has been elucidated by Andreas M. Hinz [13]. Finding diagonal sum [14], k-Fibonacci sequence, recurrence relations [15], finding exponential ( ) [16] were a part of those to describe the work that is generated from Pascal's triangle. Some fascinating properties of Pascal's triangle are available in [17,18]. In 1956, Freund [19] elicited that the generalized Pascal's triangles of th order can be constructed from the generalized binomial coefficients of order . Bankier [20] gave the Freud's alternative proof. Kallós generalized Pascal's triangle from an algebraic point of view by different bases. He tried to generalize Pascal's triangle using the power of integers [21], powers of base numbers [22] and their connections with prime numbers [23]. Kuhlmann tried to generate Pascal's triangle using the T-triangle concept [24].
The concept of using a power of 11 to generate rows of Pascal's triangle was first introduced by Sir Isaac Newton. He noticed the first five rows of Pascal's triangle are formed by a power of 11 and claimed (without proof) that subsequent rows can also be generated by a power of eleven as well [25]. Arnold et al. [26] showed if one assigns a place value to each of the individual terms in a certain row of the triangle, the pattern can be seen again. Morton [27] noted the Pascal's triangle property by the power of 11 for 10 base numeral system. Mueller [28] noted that one can get the th power of 11 from the th row of the Pascal's triangle with positional addition.
It is clearly concluded that above mentioned works did not express a full row of Pascal's triangle from a power of 11, or from the digits of (1 0 ⋯ 0 ⏟ ⏟ ⏟ proved a new general formula to generate any row of Pascal's triangle.

Methods
The very basic definition of getting the number at any cell of a row of Pascal's triangle is the summation of the numbers at the two adjacent cells of the previous row. The rows of Pascal's triangle are numbered starting from = 0 on the top and the cells in each row are numbered from = 0 on the left. For = 0, there is only one cell with the value 1. As the successive rows are generated, the numbers in the right most and left most cells are defined to be 1.  The power of 11 technique is generating Pascal's triangle by multiplying previous rows by 11 successively. The one digit partition of 11 1 = 11 gives us the numbers in the cells of the 1 st row and 11 2 = 121, 11 3 = 1331 and 11 4 = 14641 give 2 nd , 3 rd , and 4 th rows respectively. Before finding the general rule for subsequent rows, we first elaborate on the concept of powers of 11. The reason behind getting Pascal's triangle from the powers of 11 lies in the general rule of multiplication. What do we get from multiplication of a number by 11? Let be the number generated by concatenating each of the digits in the cells of the th row of Pascal's triangle from left to right.    respectively. The above scheme fails for 11 5 or 11 6 . Why does the power of 11 technique fail here, and why does the power of 11 technique work for the first four rows? If the reader closely looks at the Pascal's triangle, they will see that all of the cell values in the first to fourth rows are one digit. We get two-digit cell values for the first time in the central cells of the fifth row, which we think is a potential reason for the power of 11 technique failing here. So for finding the 5 th row onward, we need two (three, four, …) digits partitions of . The shifting of places in Fig. 2 and Fig. 3 implies using a power of 1 0 ⋯ 0 ⏟ ⏟ ⏟ Θ zeros 1, for some Θ, might work. Now, we will endeavor to formulate a specific rule.
At first, we attempt to generate the number for which two digit partitions give us the numbers in the cells of a row of the Pascal's triangle. So we extend the concept of power of 11 technique to the power of 101 technique and multiply 101 by itself to see the consequences. We can achieve this by using the very basic rules of multiplication.  Clearly, two digits partition from the right of the number 101 9 = 1093685272684360901 does not give the numbers in the cells of the 9 th row because the numbers in the central cells of this row contains three digits. So the representation of three place values for each entry of Pascal's triangle requires a new formula to be generated. The previous context directed that multiplication of a number by 11 and 101 makes the left shift of all digits by one and two places, respectively. Therefore, three-digit representation requires multiplication by 1001. Fig. 5 indicates the left shift of all digits by 3 places when a number is multiplied by 1001. By continuing the multiplication by 1001 in Fig. 4, we get from which one may form the 9 ℎ row of Pascal's triangle by three digits partition of the number from the right. But we never need a central element; rather it is necessary to know how many digits the central element has. Applying the property of the logarithmic function, one can identify how many digits (or place values) the central element has without knowing the value of the cell. Therefore, the number of digits in the central value is given by ⌈ For both of the numbers, we need 2 zeros between 1 and 1 in 11. So, to get the 9 ℎ and 10 ℎ rows we have to calculate 1001 9 and 1001 10 respectively. Both of these cases have been discussed above, and are consistent with our formula for Θ.
It's time to generate the formula to find any row of Pascal's triangle. We infer that the general formula for generating the ℎ row of Pascal's triangle is the Θ + 1 digit partitioning of the digits of the number (1 0 ⋯ 0 ⏟ ⏟ ⏟ This 16 th row can also be verified from the existing Pascal's triangle. The above formula can be used for a large . We now exemplify the 51 row of Pascal's triangle. Hence = 51 gives Θ = 14. We have to put 14 zeros between 1 and 1 in 11, that is (1000000000000001) 51 .
(1000000000000001) 51 =1000000000000051000000000001275000000000020825000000000249900000000002349060000000018009460000000115775100000 0006367630500000030423123500000127777118700000476260169700001587533899000004762601697000012927061749000031886752314200071745192706 9501477106908672502790090827492504845947226697507753515562716011445665830676015607726132740019679306863020022959191340190024795926 6474052247959266474052229591913401900196793068630200156077261327400114456658306760077535155627160048459472266975027900908274925014 7710690867250071745192706950031886752314200012927061749000004762601697000015875338990000004762601697000001277771187000000304231235 0000000636763050000000115775100000000018009460000000002349060000000000249900000000000020825000000000001275000000000000051000000000 000001 The desired 51 st row can be obtained by partitioning each 15 digits from the right. For readers' convenience, we marked each partition with different colors and showed that the above formula generates the 51 st row of the Pascal's triangle.

Results and discussion
Remark: In general the th partition of length of the digits of a positive integer is the left most digits of the number ( mod 10 × ) .
Now we give a proof of the power of 11 technique. To prove the main theorem, we prove some inequalities and lemmas. For ∈ ℕ, we have the following inequalities From the property of floor function, Θ ≤ log 10 ( Since both sides of the above inequality are integers, the difference between 10 Θ+1 and ( is at least 1, therefore From inequality (1), we also have For , ∈ ℕ and 0 ≤ ≤ , the maximum value of ( ) occurs when = ⌊ 2 ⌋. Hence ( and notice that ( From inequalities (6) and (7) ) 10 ( −2)(Θ+1) + ⋯ + 1 are the same.

Conclusion
Sir Isaac Newton hinted that binomial coefficients in the th row of the Pascal's triangle may be achieved from partitioning the digits in the th power of some number that contains 11 in some form [25]. It has been shown earlier that the weighted sum of the values in the th row of the Pascal's triangle is (11) [26]. We have proved that (Θ + 1) digit partitions of the digits of (1 0 ⋯ 0 ⏟ ⏟ ⏟ Funding statement